# Thread: Maximums through an unknown function

1. ## Maximums through an unknown function

Hi. I have a very long question which has 2 parts... I've worked out part 1 but part 2 is giving me problems.

Imagine that you are the pilot of a light aircraft which is capable of cruising at a steady speed of a km/hr in still air. You have enough fuel on board to last b hours.

You take off from the airfield and, on the outward journey, are helped along by a c km/hr wind which increases your speed relative to the ground to a + c km/hr.

Suddenly you realise on the return journey you will be flying into the wind and will therefore slow down to a-c km/hr.

What is the maximum distance that you can travel from the airfield, and still be sure that you have enough fuel left to make a safe return journey? Results must be verified.

a = 350 b = 4.5 c = 40

I found the answer to part 1 through the formula:
t = d / s by manipulating it into:
t = d / (a+c) + d / (a-c)
I subbed in the values for a, c and t and the value for a maximum one way was 777.21 km. (1554.428km there and back).

Determine a function linking the distance from the airfield (km) and the time of the flight (hrs). Hence, determine the greatest distance the plane can travel from the airfield and what wind speed will allow this to occur. You must demonstrate algebraic techniques and calculus. Conclusions must be verified.

Task 2... well, I'm honestly not quite sure where to start. I have derived the equation which I worked out to be:
t' = 2da^2 + 4dac + 2dc^2. That's all I've got so far... I could sub in the value of a (350), but that still leaves me with the unknowns of d, c and t.
Help would be much appreciated.
Thankyou.

2. Hi

For task 2 starting from $t = \frac{d}{a+c} + \frac{d}{a-c} = \frac{2ad}{a^2-c^2}$

$d = \frac{a^2-c^2}{2a}\:t$

$\frac {dd}{dc} = -\frac{c}{a}\:t$ is equal to 0 when c=0

I would say that the maximum distance is obtained when there is no wind

3. Hi,

Thanks. I will look into that.
Can I ask how you got the derivative dd/dc = -c/d * t?

4. I took the derivative of d with respect to c

$d = \frac{a^2-c^2}{2a}\:t = \frac{at}{2}-\frac{t}{2a}\:c^2$

The derivative of the first term with respect to c is equal to 0

Therefore $\frac {dd}{dc} = -\frac{t}{2a}\:2c = -\frac{c}{a}\:t$